In Δ ABC , BD ⊥ AC and CE ⊥ AB . If BD and CE intersect at O , prove that ∠ BOC = 180° - A .

Question

In &#x394; ABC , BD &#x22A5; AC and CE &#x22A5; AB . If BD and CE intersect at O , prove that &#x2220; BOC = 180&#xB0; - A .

Philoid · Accepted Answer

Given,

BD perpendicular to AC

And,

CE perpendicular to AB

In

∠E + ∠B + ∠ECB = 180^o

90^o + ∠B + ∠ECB = 180^o

∠B + ∠ECB = 90^o

∠B = 90^o - ∠ECB (i)

In

∠D + ∠C + ∠DBC = 180^o

90^o + ∠C + ∠DBC = 180^o

∠C + ∠DBC = 90^o

∠C = 90^o - ∠DBC (ii)

Adding (i) and (ii), we get

∠B + ∠C = 180^o (∠ECB + ∠DBC)

∠180^o - ∠A = 180^o (∠ECB + ∠DBC)

∠A = ∠ECB + ∠DBC

∠A = ∠OCB + ∠OBC (Therefore, ∠ECB = ∠OCB and ∠DCB = ∠OCB) (iii)

In

∠BOC + (∠OBC + ∠OCB) = 180^o

∠BOC + ∠A = 180^o [From (iii)]

∠BOC = 180^o - ∠A

Hence, proved